Math 461 B/C, Spring 2009 Midterm Exam 1 Solutions and Comments

Math 461 B/C, Spring 2009 Midterm Exam 1 Solutions and Comments 1. Suppose A, B and C are events with P (A) = P (B) = P (C) = 1/3, P (AB) = P (AC) = P (BC) = 1/4 and P (ABC) = 1/5. For each of the following events, first express the event in set-theoretic notation (e.g., (A B c ) C), then find its probability. (a) Neither A nor B occurs. Solution. The event is A c B c or (A B) c. Its probability is given by P (A c B c ) = P ((A B) c ) = 1 P (A B) = 1 P (A) P (B) + P (AB) = 1 2(1/3) + (1/4) = 7 12 (b) At least one of the three events A, B, and C occurs. Solution. The event is A B C. By inclusion/exclusion, P (A B C) = P (A) + P (B) + P (C) P (AB) P (AC) P (BC) + P (ABC) = 3 (1/3) 3 (1/4) + (1/5) = 9 20 (c) Exactly two of the three events A, B, and C occur. Solution. The event is ABC c AB c C A c BC. Now P (ABC c ) = P (AB) P (ABC) = 1/4 1/5 = 1/20, P (AB c C) = P (AC) P (ABC) = 1/4 1/5 = 1/20, P (A c BC) = P (BC) P (ABC) = 1/4 1/5 = 1/20, and since the three sets ABC c, AB c C, A c BC are disjoint, their probabilities add up, so the given event has probability P (ABC c AB c C A c BC) = 3(1/20) = 3 20 1

2. For this problem, an 8-digit number is any string of 8 decimal digits (i.e., digits 0, 1,..., 9). Leading 0 s are allowed, so that, for example, 05435199 is an 8-digit number in this sense. (a) How many 8-digit numbers consist of exactly three 1 s, three 2 s, and two 3 s? (Example: 12331221) Solution. This is a standard word-counting problem, involving three letters (1, 2, 3), with multiplicities 3, 3, and 2, respectively. The number of such words is given by the number of permutations of 8 distinct objects, namely, 8!, divided by the number of ways the repeated letters can be permuted, namely 3!, 3! and 2!: 8! 3!3!2! (= 560) (b) How many 8-digit numbers consist of exactly 4 distinct odd digits (i.e., digits 1, 3, 5, 7, 9) and 4 distinct even digits (i.e., digits 0, 2, 4, 6, 8)? (Example: 30497612 consists of the odd digits 1, 3, 7, 9 and the even digits 0, 2, 4, 6.) Solution. Method 1: First choose 4 of the 5 odd digits ( ( 5 4) choices), then choose 4 of the 5 even digits ( ( 5 4) choices), finally arrange the 8 digits chosen in some order (8! choices), to get ( )( ) 5 5 8! (= 08000) 4 4 Method 2: (This method is more complicated and not recommended.) First pick an ordered arrangement for 4 distinct odd digits (5 4 3 2 choices), then pick an ordered arrangement for 4 distinct even digits (5 4 3 2 choices), finally pick an E/O pattern consisting of 4 E s and 4 O s (8!/(4!4!) = ( 8 4) choices), to get which is the same answer as before. Comments: (5 4 3 2) 2 8! 4!4! (= 08000), (1) In the first method, the factor 8! is crucial since it accounts for the different ways the chosen digits can be ordered. (Note that the order clearly has to be taken into account: for example, 30497612 and 40397612 contain the same sets of digits, but are obviously different integers.) (2) In the second method, the factor 8!/(4!4!) is crucial, since it represents the different evenodd patterns. The factor (5 4 3 2) 2 alone only gives the number of choices that fit a fixed such pattern (e.g., EOEOEOEO or EEEEOOOO). However, in contrast to some other problems (e.g., the 5 couples problem, where men and women were required to alternate), in this problem there is no restriction on the even/odd patterns. (c) How many 8-digit numbers contain exactly 2 distinct digits? (Example: 31113131 contains exactly two distinct digits, namely 1 and 3.) Solution. [A problem of the same type was worked in class.] There are ( ) 15 pts 2 ways to choose the two digits. Once the two digits have been chosen, there are 2 8 8-digit numbers that can be formed with these digits. Of these, 2 consist of only one of the digits, leaving 2 8 2 numbers that contain both of the two digits. Thus, the total count is ( ) (2 8 2) (= 11430) 2 Comments: The factor 2 8 2 is equal to ( ) ( 8 1) + ( 8 2) + ( 8 8), as can be seen by expanding 2 8 = (1 + 1) 8 by the binomial theorem. In general, as pointed out on the cover sheet, answers 2

should not involve summations with more than a handful of terms. In the above case, full credit was given for (otherwise correct) answers containing the summation ( ), as one could regard ( ) as a handful of terms type summation (though this is a bit of a stretch). In more clear-cut cases, such answers would be penalized. Moral: If you arrive at a lengthy summation, you should look for an alternate approach that leads to a simpler answer. 3

3. For this problem assume the year has 360 days, divided into 12 months, with each month having exactly 30 days. Consider a group of people, each having one of the 360 days as birthday, with each of the 360 days being equally likely to occur as a given person s birthday. (a) What is the probability that all birthdays are distinct and fall into the same month? Solution. [This is similar to the final exam problem worked in class (which asked for the probability that all three finals fall on the same day).] Encode the possible assignments of birthdays as tuples (b 1,..., b ), with b i denoting the birthday of person i in the group, or equivalently, as a -letter birthday word b 1... b, with each b i denoting one of the 360 possible birthdays. The total number of such words, and hence the denominator in the probability formulas, is 360. Method 1: To count the number of such words in which all birthdays fall within the same month, first pick a month ( ( ) 12 1 choices), then within this month pick distinct days taking order into account (30 29 21 choices). This gives a total count of ( ) 12 1 30 29 21, and a probability 12 30 29 21 360 Method 2: First pick a birthday for the first person in the group (360 choices), then pick a birthday for the second person from the same month as that of the first birthday chosen, but different from that day (29 choices), then pick a birthday for the third person from this month, but different from the first two days chosen (28 choices), etc. This gives a total count of 360 29 28 21 and probability 360 29 28 21 360, which is the same as the above answer. Comments: (1) Using ( 30 ) instead of 30 29 21 as count for the choices of days within the month would correspond to counting unordered samples and give an answer that is off by a factor 8!. It is crucial that order is taken into account here, since otherwise the counting would be inconsistent with the denominator count, 360, which represents the total number of ordered birthday words. See the comments on Problem 28 in HW 2. (2) It is clear from the context of the problem that a random birthday assignment can contain repeated birthdays. (In this regard, the problem is different from the Final Exam Problem in class, but it is of the same type as the with replacement version of Problem 28 in HW 2.) Thus, the total number of outcomes, and hence the denominator in the probabilities, is 360, and not 360 359 351, or ( ) 360. (b) What is the probability that no two birthdays fall into the same month? Solution. [An analogous question also came up in the final exam problem worked in class.] Method 1: First pick an ordered sample of of the 12 months (12 11 3 choices), then, for each of these months chosen, pick one of the 30 days in this month (30 choices). This gives a total count of 12 11 3 30 and a probability 12 11 3 30 12 11 3 360 = 12 = 12! 2!12 Method 2: Pick a birthday for the first person in the group (360 choices), then pick a birthday for the second person from a different month than that of the first birthday chosen (330 choices), etc. This gives a total count of 360 330 300 90 and probability 360 330 300 90 360 It is easily checked that the two answers are the same. 4

15 pts (c) What is the probability that there are exactly three (distinct) pairs of matching birthdays among the people (with the remaining birthdays distinct)? (Example: 2 each have their birthday on February 19, March 19, and April 30, and 1 each has their birthday on April 1, May 11, May 14, July 4.) Solution. [This is similar to the two pairs version of the poker problem; see Problems 15(c) and 16(c) from HW 2.] First pick three distinct days for the three double birthdays ( ( ) 360 3 choices), then pick 4 distinct days from the remaining 357 days for the single birthdays ( ( ) 357 4 choices), finally arrange the resulting set of 3 double birthdays and 4 single birthdays into a birthday word (( ) 2!2!2! choices). This gives a total count of ( )( 360 357 ) 3 4! 2!2!2! and a probability ( ) ( ) 360 357! 360 359 354! 3 4 2!2!2! 360 = 3! 4! 2!2!2! 360 Comments: (1) The factor 2!2!2! (or, equivalently, ( )( 8 6 2 2)( 2) ) is crucial here since it accounts for the various ways one can arrange the selected birthdays in some order. This factor does not appear in standard poker problems since there one counts (both in #S and in #E) unordered samples, but it does appear in the poker dice problem (Problem 16 of HW 2). As explained in class (see also the comments on HW 2 and HW 3), if repetition is allowed (which is the case here since the same birthday can occur more than once), one has no choice but to use ordered samples for the counting. (2) In choosing birthdays for the three pairs, one needs to count unordered samples of 3 out of 360, so the count is ( ) 360 3, and not 360 359 358. This is a somewhat subtle issue, but it already came up in poker problems and was pointed out in class in this context. For example, ( to count poker hands with two pairs, the two values for the pairs can be chosen in 13 ) 2 ways, not 13 12; by contrast, for poker hands with one pair and one triple, the correct count is 13 12. The reason is that two A s and two B s and two B s and two A s are indistinguishable, while two A s and three B s and two B s and three A s clearly are different. (3) By the same argument, the correct count for the birthdays for the four singles is ( ) 357 4, and not 357 356 355 354. 5

4. The Putnam Contest, the World s Toughest Math Test, is a math competition for undergraduates in the U.S. and Canada. The top 15 students win prizes of $1,000 or more. In the 2006 Putnam Contest, 9 of the top 15 spots went to students from MIT, 2 went to students from Harvard, 2 went to students from Princeton, 1 went to a student from Caltech, and 1 went to a student from Waterloo. Assume there were no ties. 15 pts (a) Given the above information, how many finishing orders are possible, if only the colleges that occupy the various spots are taken into account (and not the individual students)? Solution. [Except for different numbers and different wording, this is essentially Problem 29 in Chapter 1, which was about finishing orders in a weight lifter contest.] If only the colleges are taken into account, then counting finishing orders amounts to counting 15 letter words formed with 9 M s, 2 H s, 2 P s, 1 C, and 1 W (with the M s corresponding to spots occupied by MIT students, etc.). This is a standard word counting problem, with the count is given by 15! 9!2!2! (= 900900) (b) Suppose that, in addition to the above information, it is also known that MIT occupied exactly 3 of the top 5 spots. Given this additional information, how many finishing orders (counted in the same way as above) are possible? Solution. [This is of the same type as the second part of the weight lifter problem (Problem 29 in Chapter 1), which asked for finishing orders subject to the condition that the U.S. should occupy 2 of the top 3 spots and 1 of the bottom 3 spots. Method 1: This approach is pretty straightforward, and it is the one used in the HW solutions to Problem 29. The idea is to first assign all MIT spots, then assign the remaining spots. Since MIT occupies a total of 9 spots among the top 15 and exactly 3 of the top 5, it must occupy exactly 6 of the bottom (i.e., those ranked 6 through 15). There are ( 5 3) ways to pick the 3 spots MIT occupies among the top 5, and ( ) 6 ways to pick the 6 spots MIT occupies among the bottom, so the total number of ways MIT can be assigned spots is ( 5 ) 3)( 6. Once those spots are picked, there are 6 spots left, to be filled by 2 H s, 2 P s, 1 C, and 1 W. The number of ways to fill these spots is 6!/(2!2!), by the usual word counting argument. Hence the total count is ( 5 3 )( 6 ) 6! 2!2! (= 378000) Note: A subtle, but essential, point was that in the given scenario exactly 3 of the MIT students placed in the top 5. If one overlooks this and, after assigning 3 of the top 5 slots to MIT students, treats all remaining 12 slots the same and assigns them among the remaining students (6 MIT, 2 Harvard, 2 Princeton, 1 Waterloo, 1 Caltech), one would end up with ( ) 5 12! 3 6!2!2!. This is not correct, as it would include cases where MIT has 4 or 5 of the top 5 spots. Method 2: This approach is much more complicated and fraught with pitfalls, and definitely not recommended. It was attempted by a few students, but almost none succeeded. It is mentioned here mainly to point out the many traps and pitfalls that one is likely to make if one tries that approach. 1. Start with the 9 MIT students and pick three of these for the 3 MIT spots among the top 5. This can be done in ( 9 3) ways, but (this is the first of several pitfalls) that count implies that one considers these 9 students distiguishable, contrary to the assumptions in the problem. If we ignore that issue for now and pretend the MIT students are distinguishable (e.g., labeled M 1, M 2,..., M 9 ), one can then proceed as follows: 6

2. Pick 3 of the top 5 spots for these 3 MIT students. This can be done in 5 4 3 = 5!/2! ways. (It is not ( 5 3), since to be consistent with our temporary assumption that the M s are distinguishable, we have to take order into account.) 3. Similarly, pick 6 of the bottom spots for remaining 6 MIT students. This can be done in 9 5 =!/4! ways (not ( ) 6 for the same reason as before). 4. Then, as in the first method, fill the remaining 6 slots with with 2 H s, 2 P s, 1 C, and 1 W. ( 6! 2!2! choices). 5. Up until this point the 9 M s have been considered distinguishable (e.g., labeled M 1,..., M 9 ). To undo this artificial labeling and consider all M s distingishable, divide the total count by the number of ways M 1,..., M 9 can be permuted, i.e., by 9!. Putting all these counts together, one gets ( ) 1 9 5! 9! 3 2!! 4! This does not look at all like the answer obtained above, but writing the binomial coefficients in terms of factorials, one can convert this expression to the one above: = 1 ( ) ( ) 9! 9! 5 3!6! 6! 3! 6! 3 6 2!2! ( )( ) 5 6! = 3 6 2!2! 6! 2!2! 7

15 pts 5. (a) State the three Kolmogorov axioms. Be sure to use proper mathematical notation and state any conditions that are present in the axioms. Solution. [This was a question of the state formula/theorem/property variety; the requested formulas and statements (Kolmogorov axioms and binomial theorem) were explicitly mentioned in the exam syllabus.] Axiom 1. 0 P (A) 1 for any event A. Axiom 2. P (S) = 1. Axiom 3. If A 1, A 2,... are finitely or countably infinitely many events that are mutually disjoint, then P (A 1 A 2 ) = P (A 1 ) + P (A 2 ) +. Comments: The conditions given in italics, and most importantly, the mutual disjointness, are an essential part of the third axiom. Without these conditions, the axioms would only be satisfied in trivial situations and hence would be useless. For example, without the mutual disjointness condition, one could apply (3) with A 1 and A 2 both equal to an arbitrary event A, yielding P (A) = P (A A) = P (A) + P (A) and P (A) = 0 for any event A. (b) State the Binomial Theorem. (Just state the result, making sure to use mathematically correct notation; no proofs required.) Solution. [This was one of the three bullet points in the exam syllabus for Chapter 1.] The Binomial Theorem says that, for any x, y and any positive integer n, (x + y) n = n k=0 ( ) n x k y n k k Maximal Total Score: 150 points 8